HiI'd like to ask a question I couldn't solve Hope if any one knows the answer...

A linear DNA labeled with 32p at both ends is completed digested by Bg/II to yield five fragments sizes of 3.7, 6.3, 7.0, 8.8 and 9.6 kb. Autoradiograph from southern hybridization showed that fragments 7.0 and 9.6 are radioactive, HindIII cuts the DNA to three fragments of sizes 9.2, 11.8 and 14.4 kb. Fragments 9.2 and 14.4 are radioactive. When the DNA is double digested by Bg/II and HindIII, seven fragments sized 1.5, 2.2, 4.0, 4.8, 6.3, 7.0 and 9.6 kb resulted.

a) Draw the physical map of the DNA showing the Bg/II and HindIII sites.

b) A radio-labeled gene x probe is added to the Southern blots that contain the DNA. Resulting autoradiograph showed hybridisation with 3.7 kb Bg/II and 11.8 kb HindIII fragments. Indicate the location of gene x on the physical map of the DNA.

a) this is just about counting the pieces together. I guess in the last example, again 7 and 9.6 kbp are radioactive, right?So, you know, that let's say from left you have 7 kbp, than BglII cut site. Now you have either 9.2 - 7 or 14.4 - 7 kbp to the HindIII cut site. In the second case, there would be another BglII cut sites in between, but I guess, the first case is true, right? (yeah, you have 2.2 kbp fragment in double digest) and by this way try to go futher and from both directions

b) depends on previous result. When will you have the map, just place the gene on 3.7 kbp BglII fragment AND 11.8 kbp HindIII fragment (it will be somewhere, where these fragments overlap;). This will also help you, if these fragment won't overlap, you have something wrong

c) I'm not sure, whether it is from previous or not, but doesn't matter so much. Just use appropriate primers for PCR and clone into some vector, introduce into some bacteria, grow and isolate